AN ELEMENTARY PROOF OF THE SPECTRAL RADIUS FORMULA FOR MATRICES JOEL A. TROPP Abstract. We present an elementary proof that the spectral ra- dius of a matrix A may be obtained using the formula 1 ρ(A) = lim kAnk =n; n!1 where k · k represents any matrix norm. 1. Introduction It is a well-known fact from the theory of Banach algebras that the spectral radius of any element A is given by the formula ρ(A) = lim kAnk1=n: (1.1) n!1 For a matrix, the spectrum is just the collection of eigenvalues, so this formula yields a technique for estimating for the top eigenvalue. The proof of Equation 1.1 is beautiful but advanced. See, for exam- ple, Rudin's treatment in his Functional Analysis. It turns out that elementary techniques suffice to develop the formula for matrices. 2. Preliminaries For completeness, we shall briefly introduce the major concepts re- quired in the proof. It is expected that the reader is already familiar with these ideas. 2.1. Norms. A norm is a mapping k · k from a vector space X into the nonnegative real numbers R+ which has three properties: (1) kxk = 0 if and only if x = 0; (2) kαxk = jαj kxk for any scalar α and vector x; and (3) kx + yk ≤ kxk + kyk for any vectors x and y. The most fundamental example of a norm is the Euclidean norm k·k2 which corresponds to the standard topology on Rn. It is defined by 2 2 2 kxk2 = k(x1; x2; : : : ; xn)k2 = x1 + x2 + · · · + xn: Date: 30 November 2001. q 1 SPECTRAL RADIUS FORMULA 2 One particular norm we shall consider is the `1 norm which is defined for x in Rn by the formula kxk1 = k(x1; x2; : : : ; xn)k1 = maxfxig: i For a matrix A, define the infinity norm as kAk1 = max jAijj: i j This norm is consistent with itself andPwith the `1 vector norm. That is, kABk1 ≤ kAk1kBk1 and kAxk1 ≤ kAk1kxk1; where A and B are matrices and x is a vector. Two norms k · k and k · k? on a vector space X are said be equivalent if there are positive constants C and C¯ such that ¯ Ckxk ≤ kxk ≤ C¯kxk ¯ ? for every vector x. For finite-dimensional spaces, we have the following powerful result. Theorem 2.1. All norms on a finite-dimensional vector space are equivalent. Proof. We shall demonstrate that any norm k · k on Rn is equivalent n to the Euclidean norm. Let feig be the canonical basis for R , so any vector has an expression as x = xiei. First, let us check that k · k is continuous with respect to the Euclidean norm. For all pairs of vectors x and y, P kx − yk = k (xi − yi)eik ≤ Pjxi − yij keik ≤ maxP ifkeikg jxi − yij 2 1=2 ≤ M f (xi −Pyi) g = MkxP− yk2; where M = maxifkeikg. In other words, when two vectors are nearby with respect to the Euclidean norm, they are also nearby with respect to any other norm. Notice that the Cauchy-Schwarz inequality for real numbers has played a starring role at this stage. Now, consider the unit sphere with respect to the Euclidean norm, n S = fx 2 R : kxk2 = 1g. This set is evidently closed and bounded in the Euclidean topology, so the Heine-Borel theorem shows that it SPECTRAL RADIUS FORMULA 3 is compact. Therefore, the continuous function k · k attains maximum and minimum values on S, say C and C¯. That is, ¯ Ckxk ≤ kxk ≤ C¯kxk ¯ 2 2 for any x with unit Euclidean norm. But every vector y can be ex- pressed as y = αx for some x on the Euclidean unit sphere. If we multiply the foregoing inequality by jαj and draw the scalar into the norms, we reach Ckyk ≤ kyk ≤ C¯kyk ¯ 2 2 for any vector y. It remains to check that the constants C and C¯ are positive. They are clearly nonnegative since k · k is nonnegativ¯ e, and C ≤ C¯ by defi- ¯ nition. Assume that C = 0, which implies the existence of a point x on the Euclidean unit¯sphere for which kxk = 0. But then x = 0, a contradiction. 2.2. The spectrum of a matrix. For an n-dimensional matrix A, consider the equation Ax = λx, (2.1) where x is a nonzero vector and λ is a complex number. Numbers λ which satisfy Equation 2.1 are called eigenvalues and the corresponding x are called eigenvectors. Nonzero vector solutions to this equation exist if and only if det(A − λI) = 0; (2.2) where I is the identity matrix. The left-hand side of Equation 2.2 is called the characteristic polynomial of A because it is a polynomial in λ of degree n whose solutions are identical with the eigenvalues of A. The algebraic multiplicity of an eigenvalue λ is the multiplicity of λ as a root of the characteristic polynomial. Meanwhile, the geometric multiplicity of λ is the number of linearly independent eigenvectors corresponding to this eigenvalue. The geometric multiplicity of an eigenvalue never exceeds the algebraic multiplicity. Now, the collection of eigenvalues of a matrix, along with their geometric and algebraic multiplicities, com- pletely determines the eigenstructure of the matrix. It turns out that all matrices with the same eigenstructure are similar to each other. That is, if A and B have the same eigenstructure, there exists a nonsingular matrix S such that S−1AS = B. We call the set of all eigenvalues of a matrix A its spectrum, which is written as σ(A). The spectral radius ρ(A) is defined by ρ(A) = supfjλj : λ 2 σ(A)g: SPECTRAL RADIUS FORMULA 4 In other words, the spectral radius measures the largest magnitude attained by any eigenvalue. 2.3. Jordan canonical form. We say that a matrix is in Jordan canonical form if it is block-diagonal and each block has the form d×d λ 1 λ 1 2 : : : : 3 : 6 λ 17 6 7 6 λ7 6 7 4 5 It can be shown that the lone eigenvalue of this Jordan block is λ. Moreover, the geometric multiplicity of λ is exactly one and the alge- braic multiplicity of λ is exactly d, the block size. The eigenvalues of a block-diagonal matrix are simply the eigenvalues of its blocks with the algebraic and geometric multiplicities of identical eigenvalues summed across the blocks. Therefore, a diagonal matrix composed of Jordan blocks has its eigenstructure laid bare. Using the foregoing facts, it is easy to construct a matrix in Jordan canonical form which has any eigenstructure whatsoever. Therefore, every matrix is similar to a ma- trix in Jordan canonical form. n Define the choose function k according to the following convention: n n! when k = 0; : : : ; n and = k!(n−k)! k 0 otherwise: Lemma 2.1. If J is a Jordan block with eigenvalue λ, then the com- ponents of its nth power satisfy n (J n) = λn−j+i: (2.3) ij j − i Proof. For n = 1, it is straightforward to verify that Equation 2.3 yields d×d λ 1 λ 1 2 3 J 1 = : : : : 6 λ 17 6 7 6 λ7 6 7 4 5 SPECTRAL RADIUS FORMULA 5 Now, for arbitrary indices i and j, use induction to calculate that d J n+1 J n J ij = ( )ik kj k=1 X d n 1 = λn−k+i λ1−j+k k − i j − k Xk=1 n n = λ(n−j+i)+1 + λn−j+(i+1) j − i j − (i + 1) n + 1 = λ(n+1)−j+i j − i as advertised. 3. The spectral radius formula First, we prove the following special case. Theorem 3.1. For any matrix A, the spectral radius formula holds for the infinity matrix norm: n 1=n kA k1 −! ρ(A): Proof. Throughout this argument, we shall denote the `1 vector and matrix norms by k · k. Let S be a similarity transform such that S−1AS has Jordan form: J1 −1 . J = S AS = .. : 2 3 Js 4 5 Using the consistency of k · k, we develop the following bounds. kAnk1=n = kSJ nS−1k1=n 1=n ≤ kSkkS−1k kJ nk1=n; and kS−1kkSJ nS−1kkSk 1=n kAnk1=n = kSkkS−1k −1=n ≥ kSkkS−1k kJ nk1=n: In each inequality, the former term on the right-hand side tends toward one as n approaces infinity. Therefore, we need only investigate the behavior of kJ nk1=n. SPECTRAL RADIUS FORMULA 6 Now, the matrix J is block-diagonal, so its powers are also block- diagonal with the blocks exponentiated individually: n J1 n . J = .. : 2 n3 Js 4 5 Since we are using the infinity norm, n n kJ k = max fkJk kg : k The nth root is monotonic, so we may draw it inside the maximum to obtain n 1=n n 1=n kJ k = max kJk k : k What is the norm of an exponentiated Jordan block? Recall the fact that the infinity norm of a matrix equals the greatest absolute row sum, n and apply it to the explicit form of Jk provided in Lemma 2.1. dk n n kJk k = j(Jk )1jj j=1 X dk n = jλ jn−j+1 j − 1 k j=1 X n = jλ jn jλ j1−dk d jλ jdk−j ; k k k j − 1 k ( j=1 ) X where λk is the eigenvalue of block Jk and dk is the block size.